## Precalculus (6th Edition) Blitzer

$\sin 2t-\cot t=-\cot t\cos 2t$ Recall the Trigonometric Identities and apply below, \begin{align} & \cot t=\frac{\cos t}{\sin t} \\ & \cos 2t=1-2{{\sin }^{2}}t \\ & \sin 2t=2\sin t\cos t \\ \end{align} Consider the right side of the given expression and apply the above formula, \begin{align} & -\cot t\cos 2t=-\frac{\cos t}{\sin t}\left( 1-2{{\sin }^{2}}t \right) \\ & =-\frac{\cos t}{\sin t}+\frac{\cos t2{{\sin }^{2}}t}{\sin t} \\ & =-\frac{\sin t}{\cos t}+2\sin t\cos t \\ & =-\cot t+\sin 2t \end{align} Hence, it is proved that the given identity $\sin 2t-\cot t=-\cot t\cos 2t$ holds true.